Question

The path of a rocket follows the quadratic model: h = -162 +100t + 50, where h is

the height in feet given t seconds from launch. Find out how long, to the nearest
hundredth of a second, it takes the rocket to hit the ground. (if the rocket is on the
ground, it's height, h, is now zero, right ?! )

Answer 1

Answer: 6.72 seconds

This value is approximate.

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Step-by-step explanation:

You are correct in noticing that h = 0 corresponds when the rocket is on the ground.

Plug this h value into the equation to get
`0 = -16t^2+100t+50`

You'll need the quadratic formula to solve for t.

Plug in a = -16, b = 100, c = 50.

`t = (-b\pm√(b^2-4ac))/(2a)\\\\t = (-(100)\pm√((100)^2-4(-16)(50)))/(2(-16))\\\\t = (-100\pm√(13200))/(-32)\\\\t = (-100+√(13200))/(-32) \ \text{ or } \ t = (-100-√(13200))/(-32)\\\\t \approx -0.47 \ \text{ or } \ t \approx 6.72\\\\`

Ignore the negative solution because a negative time value does not make sense. The only reasonable solution is roughly t = 6.72

It takes approximately 6.72 seconds for the rocket to hit the ground.